In two separate set-ups of Young’s double-slit experiment, fringes of equal width are observed when lights of wavelengths in the ratio 1 : 2 are used. If the ratio of the slit separations in the two cases is 2 : 1, then the ratio of the distances between the plane of the slits and the screen in the two set-ups is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4 : 1
In Young’s double-slit experiment the fringe width is \( \beta=\lambda D/d \), where \( D \) is the distance from the slits to the screen and \( d \) the slit separation. Here \( \lambda_2/\lambda_1=2 \) and \( d_1/d_2=2 \).
For equal fringe widths:
\( \displaystyle \frac{\lambda_1D_1}{d_1}=\frac{\lambda_2D_2}{d_2} \)
\( \begin{aligned} \frac{D_1}{D_2} &= \frac{\lambda_2}{\lambda_1}\times\frac{d_1}{d_2} \\ &= 2\times2 \\ &= 4 \end{aligned} \)
Therefore, \( D_1:D_2 \) is 4 : 1.